Isotopic Embeddings on Topological Spaces

# Isotopic Embeddings on Topological Spaces

 Definition: Let $X$ and $Y$ be topological spaces and let $f, g : X \to Y$ be embeddings. Then $f$ and $g$ are said to be Isotopic if there exists a continuous function $H : X \times I \to Y$ such that: 1) $H_t : X \to Y$ is an embedding for every $t \in I$. 2) $H_0 = f$. 3) $H_1 = g$. If such a function $H$ exists, then $H$ is said to be an Isotopy from $f$ to $g$.

Recall that if $X$ and $Y$ are topological spaces then a function $f : X \to Y$ is said to be an embedding if $f$ is a continuous injective function such that $f : X \to f(X)$ is a homeomorphism.

In the definition above, $I = [0, 1]$ is the closed unit interval, and the functions $H_t : X \to Y$ defined for each fixed $t \in I$ by $H_t(x) = H(x, t)$.

 Theorem 1: Let $X$ and $Y$ be topological spaces and let $f, g, h : X \to Y$ be embeddings. Then a) $f$ is isotopic to $f$. (Reflexivity) b) If $f$ is isotopic to $g$ then $g$ is isotopic to $f$. (Symmetry) c) If $f$ is isotopic to $g$ and $g$ is isotopic to $h$ then $f$ is isotopic to $h$. (Transitivity). Therefore, "to be isotopic" is an equivalence relation on the set of embeddings from $X$ to $Y$.
• Proof of a) Consider the function $H : X \times I \to Y$ defined by:
(1)
\begin{align} \quad H(x, t) = f(x) \end{align}
• Then $H$ is a continuous function since $f$ is a continuous function. Furthermore, $H_t(x) = f(x)$ is an embedding for every $t \in I$, $H_0 = f$ and $H_1 = f$. So $f$ is isotopic to $f$. $\blacksquare$
• Proof of b) Suppose that $f$ is isotopic to $g$. Then there exists a continuous function $H : X \times I \to Y$ such that $H_t$ is an embedding for every $t \in I$, $H_0 = f$, and $H_1 = g$. Consider the function $H' : X \times I \to Y$ defined by:
(2)
\begin{align} \quad H'(x, t) = H(x, 1 - t) \end{align}
• Then $H'$ is a continuous function since $H$ is a continuous function. Furthermore, $H'_t$ is an embedding for all $t \in I$, $H'_0 = H_1 = g$ and $H'_1 = H_0 = f$. So $g$ is isotopic to $f$. $\blacksquare$
• Proof of c) Suppose that $f$ is isotopic to $g$ and $g$ is isotopic to $h$. Then there exists continuous functions $H' : X \times I \to Y$ and $H'' : X \times I \to Y$ such that $H'_t$ and $H''_t$ are embeddings for all $t \in I$, $H'_0 = f$, $H'_1 = g$, $H''_0 = g$, and $H''_1 = h$. Consider the function $H : X \times I \to Y$ defined by:
(3)
\begin{align} \quad H(x, t) = \left\{\begin{matrix} H'(x, 2t) & 0 \leq t \leq \frac{1}{2}\\ H''(x, 2t - 1) & \frac{1}{2} \leq t \leq 1 \end{matrix}\right. \end{align}
• Then $H$ is a continuous function since $H'$ and $H''$ are continuous and by The Gluing Lemma. Furthermore, $H_t$ is an embedding for all $t \in I$, $H_0 = H'_0 = f$, and $H_1 = H''_1 = h$. So $f$ is isotopic to $h$. $\blacksquare$

In a sense, if we have two embeddings $f, g : X \to Y$, then $f$ and $g$ are isotopic if we can find a continuous function which "transforms" $f$ to $g$ with all the intermediate steps being embeddings from $X$ to $Y$ as well.